Question

Tìm x

a/3x²+12x=0

b/5x.(-x)²+1=6

Answer 1

b) 5x.(-x)² + 1 = 6

=> 5.x.x² = 6 - 1

=> 5.x³ = 5

=> x³ = 5:5

=> x³ = 1

=> x = 1

Answer 2

a) 3x² + 12x = 0

=> 3x(x + 4) = 0

=> x(x + 4) = 0

=> x = 0 hoặc x + 4 = 0 

+) x = 0

+) x + 4 = 0 => x = -4

Vậy: x \(\in\){0;-4}

Answer 3

a) 3x.(x+4) =0 <=> 3x = 0 --> x =0

hoặc (x+4) = 0 --> x = 4

b) 5x.(-x)² +1 = 6 <=> 5.x³ = 5 <=> x³ = 1 --> x = 1